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Mathematics
Problem
Statement-1 Let
Δ
r
=
∣
(
r
−
1
)
!
n
!
(
r
−
1
)
2
(
n
!
)
2
(
r
−
1
)
3
3
r
2
−
2
n
∣
,
then
∏
r
=
2
n
+
1
Δ
r
=
0.
\Delta_r = \begin{vmatrix} (r-1)! & n! \\ (r-1)^2 & (n!)^2 \\ (r-1)^3 & 3r^2 - 2n \end{vmatrix}, \text{ then } \prod_{r=2}^{n+1} \Delta_r = 0.
Δ
r
=
(
r
−
1
)!
(
r
−
1
)
2
(
r
−
1
)
3
n
!
(
n
!
)
2
3
r
2
−
2
n
,
then
r
=
2
∏
n
+
1
Δ
r
=
0.
Statement-2
∏
r
=
1
n
+
1
Δ
r
=
Δ
2
⋅
Δ
3
⋅
Δ
4
⋯
Δ
n
+
1
\prod_{r=1}^{n+1} \Delta_r = \Delta_2 \cdot \Delta_3 \cdot \Delta_4 \cdots \Delta_{n+1}
∏
r
=
1
n
+
1
Δ
r
=
Δ
2
⋅
Δ
3
⋅
Δ
4
⋯
Δ
n
+
1
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